How to Solve the following Cauchy Problem? I want to find $x,y: \mathbb{R} \rightarrow \mathbb{R}$ such that:
$x'(t)= -3x(t)+4y(t)$
$y'(t) = -x(t) + y(t)$
with the initial conditions $x(0) = y(0) = 1$
$\begin{bmatrix}
         x' \\
         y'
      \end{bmatrix} =  \begin{bmatrix}
         -3 & 4  \\
         -1 & 1
      \end{bmatrix} \times  \begin{bmatrix}
         x \\
         y
      \end{bmatrix}  $
Let the $A=   \begin{bmatrix}
         -3 & 4  \\
         -1 & 1
      \end{bmatrix}$
$det(A-\lambda I) = (1+\lambda)^2 = 0 \implies \lambda_1 = \lambda_2 = -1$
The eigenvectors are $V_{\lambda_1} = V_{\lambda_2} =  \begin{bmatrix}
         1  \\
         1
      \end{bmatrix}$
From there I am stuck and I don't know what to do
 A: Using Laplace Transform, we can say that:


*

*$$x'(t)=-3x(t)+4y(t)\Longleftrightarrow$$
$$\mathcal{L}_t\left[x'(t)\right]_{(s)}=\mathcal{L}_t\left[-3x(t)+4y(t)\right]_{(s)}\Longleftrightarrow$$
$$sx(s)-x(0)=-3x(s)+4y(s)\Longleftrightarrow$$
$$sx(s)-1=-3x(s)+4y(s)\Longleftrightarrow$$
$$sx(s)+3x(s)=4y(s)+1\Longleftrightarrow$$
$$x(s)\left[s+3\right]=4y(s)+1\Longleftrightarrow$$
$$x(s)=\frac{4y(s)+1}{s+3}$$

*$$y'(t)=-x(t)+y(t)\Longleftrightarrow$$
$$\mathcal{L}_t\left[y'(t)\right]_{(s)}=\mathcal{L}_t\left[-x(t)+y(t)\right]_{(s)}\Longleftrightarrow$$
$$sy(s)-y(0)=-x(s)+y(s)\Longleftrightarrow$$
$$sy(s)-1=-x(s)+y(s)\Longleftrightarrow$$
$$sy(s)-y(s)=1-x(s)\Longleftrightarrow$$
$$y(s)\left[s-1\right]=1-x(s)\Longleftrightarrow$$
$$y(s)=\frac{1-x(s)}{s-1}$$


So, we can see that:


*

*$$x(s)=\frac{4\cdot\frac{1-x(s)}{s-1}+1}{s+3}\Longleftrightarrow x(s)=\frac{s+3}{(s+1)^2}$$

*$$y(s)=\frac{1-\frac{4y(s)+1}{s+3}}{s-1}\Longleftrightarrow y(s)=\frac{s+2}{(s+1)^2}$$


With inverse Laplace Transform, we can find:


*

*$$x(t)=(1+2t)e^{-t}$$

*$$y(t)=(1+t)e^{-t}$$

